Evaluate the integral $\int_0^{\infty}\frac{x^4e^x}{(e^x-1)^2} \, dx$

Question

How do I integrate the following: $$\int_0^{\infty}\frac{x^4e^x}{(e^x-1)^2} \, dx$$

I have tried everything from integrating by parts to simply expanding the denominator, but it always makes it much more complex. No idea how to solve this.

Answer 1

As I have once heard, there is more than one way to skin an integral. And so herein, we present an approach that relies on the Polylogarithm Functions. We will engage in a series of four integration by parts that will culminate in the answer $24\text{Li}_4(1)=24\zeta(4)$.

To that end, we begin with the integral of interest $I$, which is given by

$$I=\int_0^{\infty}\frac{x^4e^x}{(e^x-1)^2}\,dx$$

Enforcing the substitution $x\to -\log (1-x)$ yields

$$I=\int_0^1\frac{\log^4 (1-x)}{x^2}\,dx \tag 1$$

We now begin a series of four straightforward integration by parts applications.

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First, letting $u=\log^4(1-x)$ and $v=-1/x$ in $(1)$ yields

$$I=-4\int_0^1\frac{\log^3 (1-x)}{x}\,dx \tag 2$$

where we tacitly applied $\lim_{x\to 1}\left(\log^4(1-x)(1-1/x)\right)=0$ to arrive at $(2)$.

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Second, letting $u=-4\log^3(1-x)$ and $v=\log x$ in $(2)$ yields

$$I=-12\int_0^1 \log^2(1-x)\frac{\log x}{1-x}\,dx \tag 3$$

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Third, letting $u=-12\log^2(1-x)$ and $v=\text{Li}_2(1-x)$ in $(3)$ yields

$$I=-24\int_0^1 \log(1-x)\frac{\text{Li}_2(1-x)}{1-x}\,dx \tag 4$$

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Fourth, letting $u=-24\log(1-x)$ and $v=-\text{Li}_3(1-x)$ in $(4)$ yields

$$\begin{align} I&=-24\int_0^1 \frac{\text{Li}_3(1-x)}{1-x}\,dx\\\\ \tag 5 &=24\text{Li}_4(1)\\\\ &=24\zeta(4)\\\\ &=24\frac{\pi^4}{90} \end{align}$$

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Thus, we have

$$\bbox[5px,border:2px solid #C0A000]{I=\frac{4\pi^4}{15}}$$

as expected and as easy as one, two, three, four!

Answer 2

If you are familiar with the Riemann Zeta Function, and also note that the integral can be written as: $$\displaystyle\int_0^{\infty} \dfrac{x^4e^{-x}}{(1-e^{-x})^2} \: dx = \displaystyle\sum_{k=1}^{\infty} k\displaystyle\int_0^{\infty}x^4 e^{-kx} \:dx = \displaystyle\sum_{k=1}^{\infty}k \cdot \dfrac{24}{k^5} = 24 \zeta(4) = \dfrac4{15} \pi^4$$

Answer 3

This is just a sequitur to Yagna Patel's answer. To prove $\zeta(4)=\frac{\pi^4}{90}$ is equivalent to proving: $$ \sum_{n\geq 0}\frac{1}{(2n+1)^4}=\frac{\pi^4}{96},\tag{1} $$ since $\zeta(4)=\sum_{n\geq 0}\frac{1}{(2n+1)^4}+\sum_{n\geq 1}\frac{1}{(2n)^4}$ gives $\zeta(4)=\frac{16}{15}\sum_{n\geq 0}\frac{1}{(2n+1)^4}$.

If we recall that the Fourier (cosine) series of $f(x)=\frac{\pi}{2}-|x|$ over $(-\pi,\pi)$ is given by: $$ f(x) = \frac{4}{\pi}\sum_{n\geq 0}\frac{\cos((2n+1)x)}{(2n+1)^2}\tag{2} $$ Parseval's identity then gives: $$ \int_{-\pi}^{\pi}f(x)^2\,dx = \frac{16}{\pi}\sum_{n\geq 0}\frac{1}{(2n+1)^4}\tag{3}$$ hence: $$ \sum_{n\geq 0}\frac{1}{(2n+1)^4}=\frac{\pi}{8}\int_{0}^{\pi}f(x)^2\,dx = \frac{\pi}{8}\cdot\frac{\pi^3}{12}=\frac{\pi^4}{96}\tag{4}$$ as wanted.

Answer 4

I am going to evaluate the integral with $x^n$ in general.

$$ \begin{aligned} \int_{0}^{\infty} \frac{x^{n} e^{x}}{\left(e^{x}-1\right)^{2}} d x =&-\int_{0}^{\infty} x^{n} d\left(\frac{1}{e^{x}-1}\right) \\ =&-\left[\frac{x^{n}}{e^{x}-1}\right]_{0}^{\infty}+n \int_{0}^{\infty} \frac{x^{n-1}}{e^{x}-1} d x \\ =& n \int_{0}^{\infty} \frac{x^{n-1} e^{-x}}{1-e^{-x}} d x \\ =& n \sum_{k=0}^{\infty} \int_{0}^{\infty} x^{n-1} e^{-(k+1) x} d x \\ =& n \sum_{k=0}^{\infty} \int_{0}^{\infty} x^{n-1} d\left(\frac{e^{-(k+1) x}}{-(k+1)}\right) \\ =& n \sum_{k=0}^{\infty}\left[-\frac{x^{n} e^{-(k+1) x}}{k+1}\right]_{0}^{\infty}+\frac{n(n-1)}{k+1} \sum_{k=0}^{\infty} \int_{0}^{\infty} x^{n-2} e^{-(k+1) x} d x \end{aligned} $$

Keeping on the integration by parts, we get $$ \begin{aligned} \int_{0}^{\infty} \frac{x^{n} e^{x}}{\left(e^{x}-1\right)^{2}} d x &=n ! \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n}} =n ! \zeta(n) \end{aligned} $$ Back to our integral, putting $n=4$ yields

$$\int_{0}^{\infty} \frac{x^{4} e^{x}}{\left(e^{x}-1\right)^{2}} d x =4 ! \sum_{k=0}^{\infty} \frac{1}{(k+1)^{4}} =4 ! \zeta(4)=\frac{4 \pi^{4}}{15} $$